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The given problem asks for the arc length function \( s(t) \) of the vector-valued curve:

\[
\mathbf{r}(t) = \left( e^{t} \cos(-5t), e^{t} \sin(-5t), e^{t} \right)
\]

with the initial point \( t = 0 \).

### To compute the arc length:
The formula for the arc length \( s(t) \) of a curve \( \mathbf{r}(t) \) is:

\[
s(t) = \int_0^t \left| \mathbf{r}'(u) \right| \, du
\]

where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \), and \( \left| \mathbf{r}'(t) \right| \) is the magnitude of the derivative.

### Step 1: Differentiate \( \mathbf{r}(t) \)
Let's differentiate each component of the curve \( \mathbf{r}(t) \):

1. \( \frac{d}{dt} \left( e^{t} \cos(-5t) \right) \)
   Using the product rule:
   \[
   \frac{d}{dt} \left( e^{t} \cos(-5t) \right) = e^{t} \cos(-5t) + e^{t} \cdot (-5) \sin(-5t)
   \]
   Simplifying:
   \[
   = e^{t} \left( \cos(-5t) + 5 \sin(-5t) \right)
   \]

2. \( \frac{d}{dt} \left( e^{t} \sin(-5t) \right) \)
   Again, applying the product rule:
   \[
   \frac{d}{dt} \left( e^{t} \sin(-5t) \right) = e^{t} \sin(-5t) + e^{t} \cdot (-5) \cos(-5t)
   \]
   Simplifying:
   \[
   = e^{t} \left( \sin(-5t) - 5 \cos(-5t) \right)
   \]

3. \( \frac{d}{dt} \left( e^{t} \right) = e^{t} \)

### Step 2: Compute \( \left| \mathbf{r}'(t) \right| \)
Now, we calculate the magnitude of \( \mathbf{r}'(t) \):

\[
\left| \mathbf{r}'(t) \right| = \sqrt{\left( e^{t} \left( \cos(-5t) + 5 \sin(-5t) \right) \right)^2 + \left( e^{t} \left( \sin(-5t) - 5 \cos(-5t) \right) \right)^2 + \left( e^{t} \right)^2}
\]

### Step 3: Simplify the magnitude
Expanding each term:

1. \( \left( e^{t} \left( \cos(-5t) + 5 \sin(-5t) \right) \right)^2 \)
   Expands to:
   \[
   e^{2t} \left( \cos^2(-5t) + 10 \cos(-5t) \sin(-5t) + 25 \sin^2(-5t) \right)
   \]

2. \( \left( e^{t} \left( \sin(-5t) - 5 \cos(-5t) \right) \right)^2 \)
   Expands to:
   \[
   e^{2t} \left( \sin^2(-5t) - 10 \cos(-5t) \sin(-5t) + 25 \cos^2(-5t) \right)
   \]

Adding these two gives:

\[
e^{2t} \left( \cos^2(-5t) + 25 \sin^2(-5t) + \sin^2(-5t) + 25 \cos^2(-5t) \right) = e^{2t} (26)
\]

Now, including the third term \( \left( e^{t} \right)^2 = e^{2t} \), the magnitude simplifies to:

\[
\left| \mathbf{r}'(t) \right| = \sqrt{e^{2t} (26 + 1)} = \sqrt{27} e^{t}
\]

### Step 4: Compute the arc length function
Now, we compute the arc length by integrating:

\[
s(t) = \int_0^t \sqrt{27} e^{u} \, du
\]

This integral is straightforward:

\[
s(t) = \sqrt{27} \left[ e^{u} \right]_0^t = \sqrt{27} \left( e^{t} - 1 \right)
\]

Thus, the arc length function is:

\[
s(t) = \sqrt{27} \left( e^{t} - 1 \right)
\]

Let me know if you'd like further clarifications or details!

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Here are some related questions to explore:

1. How is the arc length formula derived from parametric curves?
2. What is the significance of the magnitude \( \left| \mathbf{r}'(t) \right| \) in the arc length computation?
3. Can this arc length formula be applied to curves in more than three dimensions?
4. How does the product rule help in differentiating vector-valued functions?
5. How would the arc length calculation change if the curve were parameterized differently?

**Tip**: When computing arc length, always ensure that you simplify the magnitude of the derivative as much as possible before integrating.

Consider the curve r = (e^t cos(-5t), e^t sin(-5t), e^t). Compute the arc length function s(t) with the initial point t = 0.

This problem involves calculating the arc length function s(t) for the curve r = (e^t cos(-5t), e^t sin(-5t), e^t) using the fundamental theorem of calculus and vector differentiation. A step-by-step solution is provided, including key calculus formulas.

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